Microscopic body that carries the GENES for
color and other traits. One chromosome can carry hundreds of genes.

Males have two X chromosomes, females have
an X and a Y.

DOMINANT

A trait which rules over others.

The normal type is dominant to almost all
mutations.

GENE

An element in a CHROMOSOME by which hereditary
traits are determined.

Lutino is a gene carried by the X chromosome.

GENOTYPE

Genetic makeup.

A Male XC X would have a Genotype of Normal
split to Cinnamon.

MUTATION

Any variance from the normal that can replicate
itself.

Pearl-Pied is a double mutation.

PHENOTYPE

The appearance of a bird.

A Male XC X would have the phenotype of Normal.

RECESSIVE

Submissive to the DOMINANT or normal; unlike
a SEX LINKED GENE, which is carried by the X CHROMOSOME

The Pied gene is recessive to normal.

SEX LINKED

Carried on the X or male CHROMOSOME

Pearl is a Sex Linked mutation.

SPLIT

Having a GENOTYPE where a genetic trait is
hidden by DOMINANT normal. NOTE: Females can never be split to a SEX LINKED
MUTATION (Cinnamon, Pearl, Yellowcheek or Lutino).

A male XC XP would be split for both Cinnamon
and Pearl. A female X Y pN is split to Pied.

Diagramming Abbreviation

Written Abbreviation

Cinnamon

C

Cin

Lutino

L

Lut

Pearl

P

Prl

Fallow

ff

Fal

Pied

pp

Pd

Recessive Silver

ss

Sil

Whiteface

ww

WF

Yellowcheek

Yc

Yck

Dominant Silver

SS

DS

Pastel Face

pa

Pas

Male Chromosome

X

(none)

Female Chromosome

Y

(none)

Split

/

/

Double Mutation

(none)

"-"

(i.e., Lutino-Pied)

When a male cockatiel's sperm fertilizes
a female cockatiel's ovum, one half of each bird's genotype, or genetic
makeup, is donated to make the future fledgling. That means the egg is
made up of half of the father's chromosomes and half the mother's. This
is true whether we are discussing sex-linked, recessive autosomal, or dominant
autosomal mutations.

The diagram we use in genetic pairing
always has the male on the left and the female on the right:

Sire: GENTOTYPE

Dam: GENOTYPE

SONS

DAUGHTERS

When figuring up possible autosomal
(non-sex-linked) combinations, we use a diagram that looks like this, and
then plug the results into the pairing diagram above:

Males have two "X" chromosomes; females
have only one. The genes causing the color mutations Cinnamon, Lutino,
Pearl, and Yellow-Cheek, are always found on the X chromosome, NEVER on
the Y. Because there is no color gene present on the Y chromosome which
might mask the sex linked color, female cockatiels will always VISUALLY
show Cinnamon, Lutino, Pearl and/or Yellow-Cheek if it is present on their
X chromosome. A female, therefore, can never be SPLIT to Cinnamon, Lutino,
Pearl or Yellow-Cheek. A good rule to remember here is "what you see is
what you get"!

Males, however, may carry one or more
of the sex linked genes on only ONE of their two X chromosomes. If they
do, the DOMINANT NORMAL will mask that trait, and the male will be considered
SPLIT to that mutation. For example, if a male carries Pearl on only one
of his X chromosomes, which would be genetically coded X XP, he is split
to pearl. If he carries Cinnamon on the other X chromosome (XC XP), he
would be split to Cinnamon AND Pearl since he only has one gene for each
color. If, however, he had Cinnamon and Pearl on the SAME X chromosome
(XCP X) he would be split to cinnamon-pearl. NOTE: It makes no difference
whether the sex linked gene is carried on the first X chromosome or the
second in the genetic code.

A male cockatiel will always donate
one of his two X chromosomes to the offspring, but a female can donate
either her X or her Y. If the mother passes along her one X chromosome,
the offspring is MALE (X X). If she gives her Y chromosome instead, the
result is a FEMALE (X Y). In sex linked mutations, it is good to remember
that males will always throw daughters of the same mutation, and mothers
will always have sons that are at least split to any mutation they carry.

Pied, Fallow, Silver,
and Whiteface are all mutations that are also recessive to the dominant
Normal Grey. What makes them different from Sex Linked mutations? They
are autosomal, or not attached to the X chromosome. This is important because
both males AND females can be split to simple recessive mutations. The
pied gene is represented with a small "p"; silver - "s"; fallow - "f",
and whiteface - "w."

In order for one of these
colors to be visual, a cockatiel must have TWO of the genes present in
its genetic makeup, or Genotype. If only ONE recessive gene is present,
the bird is considered SPLIT to that mutation, regardless of its sex.

Both parents must carry
at least ONE of the genes for their offspring to be visually that mutation.
Crossing a bird that doesn't carry a particular recessive mutation with
one that is visually that mutation will produce offspring that appear normal
but are split for that mutation. For example, a male not carrying pied
mated to a female pied will produce all split pied babies.

When you combine the male's
recessive genes with the female's, if one does not carry any genes for
that mutation it is coded as NN (denoting non-mutated or normal). Don't
confuse this with the NORMAL of Normal Grey! If the bird only carries ONE
gene for a particular simple recessive mutation, we use one small letter
depicting the mutation followed by one large "N." For example, for split
to whiteface, we would code the recessive combination "wN." Here are some
examples using the simple recessive mutation Pied:

GENETIC
CODE

DESCRIPTION

pp

Pied

pN

Split to Pied

Np

Split to Pied

Note: It doesn't matter whether the mutated gene is
coded first or last in the pair, as both will show split.

As I said previously, when a male cockatiel's
sperm is created, it contains half the chromosomes which determine that
male's hereditary traits. The ovum in the female also contains half the
chromosomes for HER hereditary traits. This is the reason fertilization
of the ovum is necessary...without the two halves joining, there would
be no paired chromosomes and therefore no cell division, etc., that eventually
forms a baby. Even though a male has TWO X chromosomes, only ONE goes into
the creation of his offspring. The other chromosome, either X or Y, comes
from the female.

The way we show this in genetics is
by using the algebraic FOIL method .... F(irst), O(uter), I(nner), L(ast).
Let's make up a mating using numbers instead of colors ...

MALE: X1 X2 FEMALE: X3 Y

The FIRST chromosome from the male
and the FIRST chromosome from the female are combined for the first possibility....

MALE: X1 X2

FEMALE: X3 Y

F

X1 X3

The OUTER two chromosomes are then
combined (that is, the first of the male's and the last of the female's)
and because there is a Y at the end, it is a female, so it is placed on
the right side of the grid....

MALE: X1 X2

FEMALE: X3 Y

O

X1 X3

X1 Y

Next, the INNER two chromosomes are
combined (the last of the male's and the first of the female's)...and because
there are two X chromosomes, it goes on the left of the grid....

MALE:
X1 X2

FEMALE:
X3 Y

X1 X3

X1 Y

I

X2 X3

Finally, the LAST two chromosomes are
combined (the last of the male's and the last of the female's)...and again,
since it is a female, it goes on the right of the grid...

MALE: X1 X2

FEMALE: X3 Y

X1 X3

X1 Y

L

X2 X3

X2 Y

Looks simple, doesn't it? Let's work
the pairing now with colors instead of numbers attached to those X chromosomes....

Male: XC XL

Female: XP Y

FIRST:

Male: XC XL

Female: XP Y

XC XP

OUTER:

Male: XC XL

Female: XP Y

XC XP

XC Y

INNER:

Male: XC XL

Female: XP Y

XC XP

XC Y

XL XP

LAST:

Male: XC XL

Female: XP Y

XC XP

XC Y

XL XP

XL Y

Now we'll interpret those genetic codes
using percentages. We take the entire possibilities of this pairing's offspring
and look at that as 100%. Since each of these genetic codes has an equal
chance of happening based upon 100 babies produced, we say that:

· 25%
will be Normal Grey males split to Cinnamon and Pearl

· 25% will
be Normal Grey males split to Lutino and Pearl

· 25% will
be Cinnamon females, and

· 25% will
be Lutino females.

This is what we call a Sex Linked nest
.... you can determine the sex of the babies by the colors produced simply
because you know a little bit now about genetics. All your males will have
the Phenotype or appearance of being Normal, while your females will be
either Cinnamon or Lutino.

Matings start to get complicated when
you start adding in the recessive possibilities to the above grid. Let's
say that the female in the above example is not only a Pearl, but a Pearl-Pied.
Well, that would be simple! As I stated before, if one bird that visually
carries a recessive mutation is mated to another that doesn't carry the
mutation at all, all the babies will be split for that mutation. To show
that using genetic calculation, we would make use of our recessive grid....

NN

pp

Np

Np

Np

Np

Using the FOIL method and bringing
one gene per parent down, you get a possible four combinations for EACH
recessive mutation. In this situation, all four of the combinations are
the same, so 100% of the babies would be split to pied.

Let's suppose instead that both the
male and the female were SPLIT to pied. The grid would then look like this....

pN

pN

As I stated before, it doesn't matter
whether the recessive gene is on the right or left side of the combination
(i.e., Np or pN), the baby is still split for that mutation, and we code
it as pN when we "plug it into" the sex linked grid. So, now our mating
is not so simple! We have a 25% chance of our babies being full pieds,
a 50% chance of them being split to pied and a 25% chance of a baby that
isn't split at all. Okay...let's combine the outcome of our sex linked
grid and our recessive grid by charting out the possibilities of mating
a Normal Grey male split to Cinnamon, Lutino and Pied with a Pearl hen
split to Pied....

Remember, one of THESE:

MALE: XC XL

FEMALE: XP Y

XC XP

XC Y

XL XP

XL Y

Can occur with each one
of THESE:

pN

pN

pp

pN

Np

NN

Which would give you THIS:

MALE: XC XL pN

FEMALE: XP Y pN

XC XP pp

XC Y pp

XC XP pN

XC Y pN

XC XP Np

XC Y Np

XC XP NN

XC Y NN

XL XP pp

XL Y pp

XL XP pN

XL Y pN

XL XP Np

XL Y Np

XL XP NN

XL Y NN

For now, we will leave in the duplicates
(pN and Np), because it will be easier to see the percentages, based on
100% of the babies.

SONS

· 12.5% would be Pied split
to Cinnamon and Pearl;

· 25% would be Normal split
to Cinnamon, Pearl, and Pied;

· 12.5% would be Normal split
to Cinnamon and Pearl;

· 12.5% would be Pied split to
Lutino and Pearl;

· 25% would be Normal split
to Lutino, Pearl, and Pied; and

· 12.5% would be Normal split
to Lutino and Pearl.

DAUGHTERS

· 12.5% would be Cinnamon-Pied;

· 25% would be Cinnamon split
to pied;

· 12.5% would be Cinnamon;

· 12.5% would be Lutino-Pied;

· 25% would be Lutino split to
pied; and

· 12.5% would be Lutino.

Calculating
Percentages

Here's a way you can figure out the
number of Genotypes you will get from any combination of sex-linked and
recessive genes.

Figure the number of combinations
for EACH recessive pairing. In this case there is only ONE possible combination
of the recessive genes: fN. There are THREE possible combinations of whiteface:
ww, wN, and NN (which we leave blank). There are also THREE possible combinations
of pied: pp, pN, and NN.

Calculate the number of combinations by
multiplying the number of possibilities:

4 sex-linked

x 1 fallow

4

x 3 whiteface

12

x 3 pied

36 possible combinations

Now let's determine the PERCENTAGE
OF OCCURRENCE. When you mate two birds that are SPLIT to a recessive mutation,
25% of the resulting offspring are FULL VISUAL mutations, 50% are split
for that mutation, and 25% carry no genetic trait for that mutation. If,
for example, you mate a split pied to a split pied, the chart looks like
this:

pN

pN

pp

pN

Np

NN

Regardless of whether the recessive
gene appears first or last in the genetic code (pN or Np), the outcome
is still the same: SPLIT. Out of the four, one is pp, or fully visual (25%);
two are split (50%); and one has no pied genes (25%). If, however, you
mate birds that are split for more than one gene, the percentages change:

pN sN

pN sN

pp ss - visual pied and silver

pp sN - visual pied split to silver

pp Ns - visual pied split to silver

pp NN - visual pied (not carrying silver)

pN ss - visual silver split to pied

pN Ns - split to silver and pied

pN NN - split to pied (not carrying silver)

Np ss - visual silver split to pied

Np sN - split to silver and pied

Np Ns - split to silver and pied

Np NN - split to pied (not carrying silver)

NN ss - visual silver (not carrying pied)

NN sN - split to silver (not carrying pied)

NN Ns - split to silver (not carrying pied)

NN NN - not carrying pied or silver

There are 16 different
ways the pairs of genes could combine.

If you are figuring percentages,
you divide 100 by the number of possibilities, and you come up with each
combination occurring 6.25% of the time! You will notice, however, that
certain combinations are coded differently but mean the same thing...for
example, pp sN and pp Ns. There is only one pp ss and only one NN NN...so
these each have a 6.25% chance of occurring. There are, however, two "visual
pied split to silver," so that genetic code has a 6.25% x 2 (or 12.5%)
chance of occurring. There are 4 "split to silver and pied"....6.25% x
4 = 25% chance of occurrence. There is only one visual pied not carrying
silver, so that chance of occurring remains 6.25%. And so on... This is,
of course, based on 100 birds produced by one pair, so if in one clutch
you get four visual silver-pieds, you are very lucky!

Now, let's compute the percentage of
the previous pair with the addition of sex-linked mutations.

MALE: XP XP pN sN

FEMALE: X Y pN sN

Here we have a male pearl
split to pied and silver mated to a female normal split to pied and silver.
Looking at just the sex-linked combinations, all the males will be normals
split to pearl, and all the females will be pearl....so this will have
no effect on the percentages, with the exception of splitting up the males
and the females. Instead of dividing the percentages among the entire offspring,
we look at the percentages separately for sons and daughters.

SONS

DAUGHTERS

6.25% Silver-Pied/Pearl

6.25% Silver-Pearl-Pied

12.50% Pied/Pearl & Silver

12.50% Pearl-Pied/Silver

6.25% Pied/Pearl

6.25% Pearl-Pied

12.50% Silver/Pearl & Pied

12.50% Silver-Pearl/Pied

25.00% Normal/Silver,Pearl
& Pied

25.00% Pearl/Silver & Pied

12.50% Normal/Pearl & Pied

12.50% Pearl/Pied

6.25% Silver/Pearl

6.25% Silver-Pearl

12.50% Normal/Silver &
Pearl

12.50% Pearl/Silver

6.25% Normal/Pearl

6.25% Pearl

You will have to adjust the percentages
when you add more sex-linked possibilities.

Let's change that male to being split
to Cinnamon and Pearl and see what happens:

MALE: XP XC pN sN

FEMALE: X Y pN sN

SONS:

XP X pp ss

DAUGHTERS

XP Y pp ss

XP X pp sN

XP Y pp sN

XP X pp NN

XP Y pp NN

XP X pN ss

XP Y pN ss

XP X pN sN

XP Y pN sN

XP X pN NN

XP Y pN NN

XP X NN ss

XP Y NN ss

XP X NN sN

XP Y NN sN

XP X NN NN

XP Y NN NN

XC X pp ss

XC Y pp ss

XC X pp sN

XC Y pp sN

XC X pp NN

XC Y pp NN

XC X pN ss

XC Y pN ss

XC X pN sN

XC Y pN sN

XC X pN NN

XC Y pN NN

XC X NN ss

XC Y NN ss

XC X NN sN

XC Y NN sN

XC X NN NN

XC Y NN NN

WHEW!

To determine the percentages, just
take the ones you figured up before, divide them in half, and substitute
CINNAMON for PEARL in the other half!

SONS

DAUGHTERS

3.125% Silver-Pied/Pearl

3.125% Silver-Pearl-Pied

6.25% Pied/Pearl & Silver

6.25% Pearl-Pied/Silver

3.125% Pied/Pearl

3.125% Pearl-Pied

6.25% Silver/Pearl & Pied

6.25% Silver-Pearl/Pied

12.50% Normal/Silver,Pearl & Pied

12.50% Pearl/Silver & Pied

6.25% Normal/Pearl & Pied

6.25% Pearl/Pied

3.125% Silver/Pearl

3.125% Silver-Pearl

6.25% Normal/Silver & Pearl

6.25% Pearl/Silver

3.125% Normal/Pearl

3.125% Pearl

3.125% Silver-Pied/Cinnamon

3.125% Silver-Cinnamon-Pied

6.25% Pied/Cinnamon & Silver

6.25% Cinnamon-Pied/Silver

3.125% Pied/Cinnamon

3.125% Cinnamon-Pied

6.25% Silver/Cinnamon & Pied

6.25% Silver-Cinnamon/Pied

12.50% Normal/Silver,Cinnamon & Pied

12.50% Cinnamon/Silver & Pied

6.25% Normal/Cinnamon & Pied

6.25% Cinnamon/Pied

3.125% Silver/Cinnamon

3.125% Silver-Cinnamon

6.25% Normal/Silver & Cinnamon

6.25% Cinnamon/Silver

3.125% Normal/Cinnamon

3.125% Cinnamon

Enough of this! When the percentages
get this small, they mean very little to you. What is important, usually,
is knowing the possibilities that CAN happen, and, if you can, the sex
of the babies.

Imagine you have two dogs, standing
side by side. One dog has two fleas, the other has none. One of the fleas
jumped from the first dog to the second dog, and now each dog has a flea
on his back!

Crossover is the "jumping" of one sex-linked
gene (the flea) from one X chromosome (dog) to another! Since you need
two X chromosomes for crossover to occur, it is obvious that it only happens
with males. There are only a few rules regarding crossover:

1. The gene cannot occur on BOTH X
chromosomes. For example, this means that a male with a genetic description
XC XC cannot experience a crossover.

2. It is not necessary for the entire
clutch to be affected by crossover. Only one bird out of a clutch, for
example, could have sex-linked genes that result from a crossover.3. Crossover occurs, on the average
with Cinnamon, Pearl and Lutino, only 25-30% of the time. (Yellowcheek
has a much lower percentage of crossover.) Let's look at crossover
a little closer, using a normal male split to cinnamon and pearl, mated
to a normal female.

MALE: XP XC

FEMALE: X Y

XP X

XP Y

XC X

XC Y

This would be the expected outcome
of the above mating...50% of the males will be split to Pearl, 50% of the
males will be split to Cinnamon; 50% of the females will be pearl and 50%
will be cinnamon. But how do you think we ever got the beautiful double
mutation Cinnamon-Pearl? It was the result of crossover.

X P X C =
X XCP

This makes
the maleís genetic coding X XCP, and the mating chart like this:

MALE: X XCP

FEMALE: X Y

X X

X Y

XCP X

XCP Y

Suddenly, he has produced Normal and
Cinnamon-Pearl female offspring. Without crossover, he could only produce
Pearl and Cinnamon hens, with all Normal males.

There is also such a thing as REVERSE
crossover, where a male that contains a double mutation on one X chromosome
passes on only a single sex-linked mutation!

The Dominant Silver mutation
is the first which is, as its name indicates, equally as dominant as the
normal Grey. When a normal cockatiel has one gene for a simple recessive
mutation, we call him "split" to that mutation. If he is, however, carrying
only one gene for Dominant Silver, he is considered a Single Factor (SF)
Silver. A SF Dominant Silver is also known as a dilute Dominant Silver,
meaning that the silver color is diluted, or washed out, by the normal
Grey. Predictably, a cockatiel with both Dominant Silver genes is called
a Double Factor (DF) Silver.

In the SF Dominant Silver
mutation, the color of the normal cockatiel becomes a silvery shade of
Grey. The male matures to carry the yellow face mask and orange cheek patches;
both males and females have a slightly darker head and neck, as well as
Grey legs and black eyes. If you recall, the recessive silver mutation
has RED eyes; the Dominant Silver retains the dark eyes of the normal.

The DF Dominant Silver
is an even further dilution of silver grey, giving an almost yellowish-white
coloration to the wings and tail feathers. The eyes continue to be black,
and the legs are, again, grey.

Inheritance of Dominant
Silver is similar to that of the simple recessive mutations. The genes
exist apart from the sex-linked chromosomes; therefore, both males and
females can have the single factor. A normal Grey SF Dominant Silver cock
would be genetically coded X X SN, while a normal Grey DF Dominant Silver
cock would be X X SS. Of course, all the mutations can be combined with
the Dominant Silver mutations.

First, letís explore the
inheritance of the Dominant Silver gene. A male Dominant Silver mated to
a female normal Grey would be coded:

Male: X X SS

Female: X Y (NN)

Sons: X X SN

Daughters: X Y
SN

All the babies from this
pair would be Single Factor Dominant Silver cockatiels, regardless of whether
they are males or females. Similarly, a female Double Factor Dominant Silver
mated to a normal Grey cock would produce all SF Dominant Silver babies
as well.

Think back to our discussion
of Simple Recessive mutations. The Dominant Silver mutation is inherited
just like the pied, whiteface, fallow, and silver recessives. The only
difference is, we code the Dominant Silver with a capital "S" to denote
its equality with the normal Grey. So, when genetically coding the inheritance
of Dominant Silver, treat it just like recessive silver (but in the upper
case)!

DOMINANT
SILVER

RECESSIVE
SILVER

SS

NN

SN

SN

NN

SN

NN

SN

SN

SS (25%)

SN (50%)

NN (25%)

SS

SN

SS

SN

SS

SS

SS

ss

NN

sN

sN

NN

sN

NN

sN

sN

ss (25%)

sN (50%)

NN (25%)

ss

sN

ss

sN

ss

ss

ss

Now, to combine the Dominant Silver
mutation with other mutations, you add one of the combinations above to
your genetic charts! For example, a SF Dominant Silver Pearl cock mated
to a DF Dominant Silver hen would be charted like this:

1. Separate the sex-linked mutations
out, and chart them - -

Male: XP XP

Female: X Y

Sons: XP X

Daughters: XP Y

2. Then chart out any simple recessive
or Dominant Silver mutations. (There are no simple recessives in this mating.)

Male: Single Factor (SN)

Female: Double Factor (SS)

Double Factor (SS) (50%)

Single Factor (SN) (50%)

3. Now, combine the two. Remember,
each of the Dominant Silver combinations can occur with each of the sex-linked
combinations!

Now,
let me start by saying THERE IS CURRENTLY NO SUCH THING AS A TRUE ALBINO
COCKATIEL! Oh, I can hear the phone ringing now...."What do you mean thereís
no such thing as an albino cockatiel? Iíve been breeding them for years!"
Before you start dialing, read on: what we call an albino cockatiel certainly
does have the PHENOTYPE (appearance) of an albino. It has all white feathers,
pink feet, red eyes, and no cheek patch. Sounds like an albino to me! But
the reason it APPEARS to be an albino is that it is really a combination
of the sex-linked mutation, Lutino, which "hides" the melanin pigment (gray,
cinnamon, silver, etc.), and the autosomal recessive mutation, whiteface,
which "hides" the lipochrome (yellow and orange) pigments. Thus, a bird
is created without pigmentation, and appears to be albino. GENETICALLY,
however, it is NOT AN ALBINO....it is a Lutino-Whiteface. It is a COMBINATION
OF TWO MUTATIONS, not its own true mutation. The so-called "albino" cannot
pass along ONE GENE to produce another albino. Instead, one sex-linked
gene and one recessive gene are passed to the offspring. If a Lutino-Whiteface
(albino) male were mated to a normal hen, the genetic chart would look
like this:

Male: XL XL ww

Female: X Y

Sons: XL X wN

Daughters: XL Y wN

All the male offspring
produced by this mating would be normals split to Lutino and Whiteface.
(By the way, donít ever make the mistake of saying "split to albino," or
Iíll have to send the genetically-correct police after you!) The female
babies would all be Lutino split to whiteface. Notice that NONE of the
babies are "albino" in appearance! If there were a TRUE albino mutation,
it would exist as one gene (probably sex-linked), and the female babies
would all look like the father. But, since the "albino" mutation is really
Lutino-Whiteface, the female babies take on all the fatherís sex-linked
genes, and have the appearance of Lutino while also inheriting the one
gene for Whiteface.

Now that I have exhausted
that subject, I will finally discuss how to produce a Lutino-Whiteface
baby. There are several ways to do this. The easiest way, of course, is
to go out and buy a Lutino-Whiteface pair! But thatís the easy way out,
and I personally donít find that too challenging. The better way is to
develop a good Lutino line and a good Whiteface line, and then to use the
best of both for your breeding of Lutino-Whitefaces.

A good Lutino male
(letís try to pick one without a bald spot, please) mated to a nice whiteface
hen (just a plain whiteface will do), will produce this:

Male: XL XL

Female: X Y ww

Sons: XL X wN

Daughters: XL Y wN

The male offspring will
all be split to both Lutino and Whiteface, and all hens will be Lutino
split to whiteface. Now, to any of you who said, "Oh, good, now I can breed
the brother to the sister to get albino babies!"....a slap on the wrist
and a mighty TSK TSK. Donít you know that inbreeding is the way we developed
those nasty bald spots to begin with?

Letís take the biggest
male produced by the above couple and mate him back to one of his female
whiteface cousins on his motherís side. That charting will look like this:

Male: XL X wN

Female: X Y ww

XL X wN

XL Y wN

XL X ww

XL Y ww

X X wN

X Y wN

X X ww

X Y ww

As you can see, 25% of
your males will be normal split to Lutino and Whiteface, 25% will be Whiteface
split to Lutino, 25% will be normal split to whiteface, and 25% will be
whiteface; 25% of your females will be Lutino split to whiteface, 25% will
be Lutino-Whiteface, 25% will be normal split to whiteface, and 25% will
be whiteface. Of course, the only babies above that will be useful in producing
further Lutino-Whiteface offspring will be those carrying both Lutino and
Whiteface! Since you canít tell which males are split to Lutino, you are
better off keeping just the female Lutinos and Lutino-Whitefaces for future
breeding.

The pastel face cockatiel is very similar
to the Yellowcheek in coloration, with a yellow cheek patch instead of
the orange of the Normal Grey. The mode of inheritance, however, is completely
different than that of the Yellowcheek.

The pastel gene is autosomal rather
than sex-linked, and is recessive to all mutations except the Whiteface.
When bred to the Normal Grey (or any mutation other than Whiteface), both
the male and female offspring are split to Pastel Face, as in the example
below:

Male: X X papa

Female: X Y

Sons: X X paN

Daughters: X Y paN

If the Male in this mating was, instead,
a Pearl-Pied-Pastel, and the female was a Pied, the genetic charting would
look like this:

Male: XP XP papa pp

Female: X Y pp

Sons: XP X paN pp

Daughters: XP Y paN pp

All those pís and Pís can get confusing!

When paired with the Whiteface, however,
the equation changes radically. It is believed that Pastel and Whiteface
reside on the same pair of autosomal chromosomes. When even one Pastel
gene exists with a matching Whiteface gene, the Pastel becomes VISUAL!
For Pastel to be visual in cockatiels WITHOUT Whiteface, however, there
must be TWO Pastel Face genes present. Elsie Burgin, one of NCSís most
highly respected breeders, gave me some good information about possible
expectations from breeding Pastels. Iíve charted them genetically for you
below. Donít be surprised if you have to read this three or four times
before you understand it completely!

Cynthia Kiesewetter
has been owned by cockatiels, as of this writing, for approximately 15 years. It was her father, James Dodson (one of the founders of the Connecticut
Budgerigar Society), who instigated her interest in aviculture. His knowledge
of budgerigar genetics was instrumental in spurring her to learn more about
cockatiel genetics.

Cynthia was
active in her local club, Connecticut Association for Aviculture, for 8
years as Secretary, Bulletin Editor, Show Advisory Committee Chairperson,
and Membership Co-chairperson with her husband, Michael. They have also been members of the American Federation of Aviculture, Society of Parrot Breeders
and Exhibitors, American Cockatiel Society, National Cockatiel Society,
and the North American Cockatiel Society.

In the National
Cockatiel Society, Cynthia (better known as Cindy to her friends) served
as Connecticut State Coordinator, Nomination Chairperson, and, from early
1993 to late 1997, Band and Membership Secretary and NCS Magazine Genetics
Consultant. She was awarded the N.C.S. Presidential Award in 1993, 1994.
1995, and 1996 for outstanding contributions. Cynthia and Michael
also created and maintained an Internet site for the
National Cockatiel Society for two years. Cynthia resigned her NCS
duties in late 1997 to form the North American Cockatiel Society and dedicate
herself more to her birds and her newly expanding family.

Copyright 2000 by Cynthia Kiesewetter and the North American Cockatiel Society.

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